Let $(x,y)$ be the local coordinates on a Riemannian manifold $M$ with $\dim(M) =2$. Let $\star$ denote the Hodge-$\star$ operator, and let $g = g_{ij}$ denote the Riemannian metric on $M$.
I am attempting to compute the formula for the Laplace–Beltrami acting on 1-forms of $M$. I have been stuck on computing that action of the Hodge-$\star$ operator on $dx$ and $dy$, and this is what I would like help with.
We know: $\star(1) = \sqrt{\det(g)} dx \wedge dy$ and $\star(dx \wedge dy) = \dfrac{1}{\sqrt{\det(g)}}$. If the metric is simply $\delta_{ij}$, then we know that $\star dx$ and $\star dy = -dx$, but I am interested in the case when the metric $g_{ij} \neq \delta_{ij}$.
I would appreciate any help with this, and if you need any further details, please let me know.
I have been aware of the following, but has been of little or no avail:\begin{eqnarray*}
dx \wedge \star dx &=& \sqrt{\det(g_{ij})} g^{11} dx \wedge dy,\\
dy \wedge \star dx &=& \sqrt{\det(g_{ij})} g^{12} dy \wedge dx
\end{eqnarray*}
From this, it is claimed that we deduce: $$\star dx = \sqrt{\det(g_{ij})}(g^{11} dy – g^{12}dx).$$
This is not clear.
Best Answer
Write $\star dx = P dx + Q dy$ for some $P,Q$ to be determined. We have $$dx \wedge \star dx = P dx \wedge dx + Q dx \wedge dy = Q dx \wedge dy,$$ and $$dy \wedge \star dx = P dy \wedge dx + Q dy \wedge dy = P dy \wedge dx.$$
With $dx \wedge \star dx = \sqrt{\det(g_{ij})} g^{11} dx \wedge dy$, we observe that $Q = \sqrt{\det(g_{ij})} g^{11}$. Similarly, with $dy \wedge \star dx = \sqrt{\det(g_{ij})} g^{12} dx \wedge dy$, we have $P = - g^{12} \sqrt{\det(g_{ij})}$. Hence, $$\star dx = \sqrt{\det(g_{ij})} g^{11} dy - \sqrt{\det(g_{ij})} g^{12} dx,$$ as claimed.